Question #98dd6

1 Answer
Nov 25, 2015

Answer:

#6 * 10^(23)#

Explanation:

You know that one mole of any substance contains exactly #6.022 * 10^(23)# atoms or molecules of that substance - this is known as Avogadro's number.

In your case, you're dealing with a millimole, which represents the #1/1000"th"# part of a mole.

So, if you need #10^3# millimoles to make one mole, it follows that you'd need #10^3# times more molecules of carbon dioxide, #"CO"_2#, to get #6.022 * 10^(23)# molecules of carbon dioxide.

#1color(red)(cancel(color(black)("mmole"))) * (1color(red)(cancel(color(black)("mole"))))/(10^3color(red)(cancel(color(black)("mmoles")))) * (6.022 * 10^(23)"molecules CO"_2)/(1color(red)(cancel(color(black)("mole")))) = 6.022 * 10^(23)"molecules CO"_2#

You need to round this off to one sig fig, the number of sig figs you have for the number of millimoles of #"CO"_2#.

#"no. of molecules of CO"_2 = color(green)(6 * 10^(23))#